\(\int \frac {\sec ^3(c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx\) [560]

Optimal result

Integrand size = 23, antiderivative size = 140 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx=-\frac {\sqrt {7} E\left (\frac {1}{2} (c+\pi +d x)|\frac {8}{7}\right )}{3 d}+\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+\pi +d x),\frac {8}{7}\right )}{3 \sqrt {7} d}-\frac {\sqrt {7} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+\pi +d x),\frac {8}{7}\right )}{3 d}+\frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac {\sqrt {3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d} \]

[Out]

-1/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)*EllipticF(cos(1/2*d*x+1/2*c),2/7*14^(1/2))/d*7^(1/2)+1/3
*(sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)*EllipticE(cos(1/2*d*x+1/2*c),2/7*14^(1/2))/d*7^(1/2)+1/3*(sin
(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)*EllipticPi(cos(1/2*d*x+1/2*c),2,2/7*14^(1/2))/d*7^(1/2)+1/3*(3-4*c
os(d*x+c))^(1/2)*tan(d*x+c)/d+1/6*sec(d*x+c)*(3-4*cos(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2881, 3134, 3138, 2733, 3081, 2741, 2885} \[ \int \frac {\sec ^3(c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx=\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{3 \sqrt {7} d}-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x+\pi )|\frac {8}{7}\right )}{3 d}-\frac {\sqrt {7} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{3 d}+\frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x) \sec (c+d x)}{6 d} \]

[In]

Int[Sec[c + d*x]^3/Sqrt[3 - 4*Cos[c + d*x]],x]

[Out]

-1/3*(Sqrt[7]*EllipticE[(c + Pi + d*x)/2, 8/7])/d + EllipticF[(c + Pi + d*x)/2, 8/7]/(3*Sqrt[7]*d) - (Sqrt[7]*
EllipticPi[2, (c + Pi + d*x)/2, 8/7])/(3*d) + (Sqrt[3 - 4*Cos[c + d*x]]*Tan[c + d*x])/(3*d) + (Sqrt[3 - 4*Cos[
c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(6*d)

Rule 2733

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a - b]/d)*EllipticE[(1/2)*(c + Pi/2
+ d*x), -2*(b/(a - b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a - b, 0]

Rule 2741

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a - b]))*EllipticF[(1/2)*(c + P
i/2 + d*x), -2*(b/(a - b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a - b, 0]

Rule 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2
- b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])
^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m +
n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||
!(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 2885

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a - b)*Sqrt[c - d]))*EllipticPi[-2*(b/(a - b)), (1/2)*(e + Pi/2 + f*x), -2*(d/(c - d))], x] /; FreeQ[{
a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c - d, 0]

Rule 3081

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3138

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d}+\frac {1}{6} \int \frac {\left (6+3 \cos (c+d x)-2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx \\ & = \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac {\sqrt {3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d}+\frac {1}{18} \int \frac {\left (21-6 \cos (c+d x)+12 \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx \\ & = \frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac {\sqrt {3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d}+\frac {1}{72} \int \frac {(84+12 \cos (c+d x)) \sec (c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx-\frac {1}{6} \int \sqrt {3-4 \cos (c+d x)} \, dx \\ & = -\frac {\sqrt {7} E\left (\frac {1}{2} (c+\pi +d x)|\frac {8}{7}\right )}{3 d}+\frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac {\sqrt {3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d}+\frac {1}{6} \int \frac {1}{\sqrt {3-4 \cos (c+d x)}} \, dx+\frac {7}{6} \int \frac {\sec (c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx \\ & = -\frac {\sqrt {7} E\left (\frac {1}{2} (c+\pi +d x)|\frac {8}{7}\right )}{3 d}+\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+\pi +d x),\frac {8}{7}\right )}{3 \sqrt {7} d}-\frac {\sqrt {7} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+\pi +d x),\frac {8}{7}\right )}{3 d}+\frac {\sqrt {3-4 \cos (c+d x)} \tan (c+d x)}{3 d}+\frac {\sqrt {3-4 \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{6 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.30 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.69 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx=\frac {-\frac {4 \sqrt {-3+4 \cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),8\right )}{\sqrt {3-4 \cos (c+d x)}}+\frac {18 \sqrt {-3+4 \cos (c+d x)} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),8\right )}{\sqrt {3-4 \cos (c+d x)}}-\frac {2 i \left (21 E\left (i \text {arcsinh}\left (\sqrt {3-4 \cos (c+d x)}\right )|-\frac {1}{7}\right )-12 \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {3-4 \cos (c+d x)}\right ),-\frac {1}{7}\right )-8 \operatorname {EllipticPi}\left (-\frac {1}{3},i \text {arcsinh}\left (\sqrt {3-4 \cos (c+d x)}\right ),-\frac {1}{7}\right )\right ) \sin (c+d x)}{3 \sqrt {7} \sqrt {\sin ^2(c+d x)}}+\sqrt {3-4 \cos (c+d x)} (1+2 \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{6 d} \]

[In]

Integrate[Sec[c + d*x]^3/Sqrt[3 - 4*Cos[c + d*x]],x]

[Out]

((-4*Sqrt[-3 + 4*Cos[c + d*x]]*EllipticF[(c + d*x)/2, 8])/Sqrt[3 - 4*Cos[c + d*x]] + (18*Sqrt[-3 + 4*Cos[c + d
*x]]*EllipticPi[2, (c + d*x)/2, 8])/Sqrt[3 - 4*Cos[c + d*x]] - (((2*I)/3)*(21*EllipticE[I*ArcSinh[Sqrt[3 - 4*C
os[c + d*x]]], -1/7] - 12*EllipticF[I*ArcSinh[Sqrt[3 - 4*Cos[c + d*x]]], -1/7] - 8*EllipticPi[-1/3, I*ArcSinh[
Sqrt[3 - 4*Cos[c + d*x]]], -1/7])*Sin[c + d*x])/(Sqrt[7]*Sqrt[Sin[c + d*x]^2]) + Sqrt[3 - 4*Cos[c + d*x]]*(1 +
2*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(6*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(407\) vs. \(2(195)=390\).

Time = 3.45 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.91

method	result	size
default	\(-\frac {\sqrt {-\left (8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}^{2}}-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {56 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 \sqrt {14}}{7}\right )}{21 \sqrt {8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {56 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 \sqrt {14}}{7}\right )}{3 \sqrt {8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {56 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \frac {2 \sqrt {14}}{7}\right )}{3 \sqrt {8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7}\, d}\)	\(408\)

[In]

int(sec(d*x+c)^3/(3-4*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-(-(8*cos(1/2*d*x+1/2*c)^2-7)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1/3*cos(1/2*d*x+1/2*c)*(8*sin(1/2*d*x+1/2*c)^4-sin
(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2-2/3*cos(1/2*d*x+1/2*c)*(8*sin(1/2*d*x+1/2*c)^4-sin(1/2*d
*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+1/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^(1
/2)/(8*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2/7*14^(1/2))-1/3*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^(1/2)/(8*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)*El
lipticE(cos(1/2*d*x+1/2*c),2/7*14^(1/2))-1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^(1/2)/(8
*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,2/7*14^(1/2)))/sin(1/2*d*x+1
/2*c)/(-8*cos(1/2*d*x+1/2*c)^2+7)^(1/2)/d

Fricas [F]

\[ \int \frac {\sec ^3(c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{\sqrt {-4 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

integrate(sec(d*x+c)^3/(3-4*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-4*cos(d*x + c) + 3)*sec(d*x + c)^3/(4*cos(d*x + c) - 3), x)

Sympy [F]

\[ \int \frac {\sec ^3(c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\sqrt {3 - 4 \cos {\left (c + d x \right )}}}\, dx \]

[In]

integrate(sec(d*x+c)**3/(3-4*cos(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**3/sqrt(3 - 4*cos(c + d*x)), x)

Maxima [F]

\[ \int \frac {\sec ^3(c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{\sqrt {-4 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

integrate(sec(d*x+c)^3/(3-4*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^3/sqrt(-4*cos(d*x + c) + 3), x)

Giac [F]

\[ \int \frac {\sec ^3(c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{\sqrt {-4 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

integrate(sec(d*x+c)^3/(3-4*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^3/sqrt(-4*cos(d*x + c) + 3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x)}{\sqrt {3-4 \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,\sqrt {3-4\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

int(1/(cos(c + d*x)^3*(3 - 4*cos(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(3 - 4*cos(c + d*x))^(1/2)), x)